zzr0ck3r
  • zzr0ck3r
Good methods for finding eigen values? \(M = \begin{array}{cc} 0&1&1&1&1& \\ 1&0&1&1&1 \\ 1&1&0&1&1 \\ 1&1&1&0&1\\ 1&1&1&1&0 \end{array}\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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zzr0ck3r
  • zzr0ck3r
@freckles where you go ;(
UsukiDoll
  • UsukiDoll
\[\det(A- \lambda I) =0\] and then use cofactor expansion like crazy
zzr0ck3r
  • zzr0ck3r
I said good methods :)

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UsukiDoll
  • UsukiDoll
=(
UsukiDoll
  • UsukiDoll
row operations first? idk that may destroy the determinant...
freckles
  • freckles
lol well ... I was kind of trying to see if could find a pattern but I didn't see a pattern like you know M_n is a n by n matrix with 0's on the diagonal and 1's everywhere else but it was totally not a good method
freckles
  • freckles
and not just a bad method but it didn't seem to work like that
UsukiDoll
  • UsukiDoll
does this whole matrix break one of the rules for row operations? Don't we need a 1 at the top left corner of the first row?
zzr0ck3r
  • zzr0ck3r
word
UsukiDoll
  • UsukiDoll
row swaps... so it does not break the first rule
UsukiDoll
  • UsukiDoll
I would switch the last one with the first one r_1<->r_5
UsukiDoll
  • UsukiDoll
swap r_2 <-> r_4 , then r_3 <-> r_4 ?
freckles
  • freckles
actually I do see a pattern
freckles
  • freckles
so there must be some kinda of shortcut for M_n where M_n is a n by n matrix with 0's a long the diagonal and 1's elsewhere
freckles
  • freckles
I want to make a conjecture \[\text{ egienvalues }(M_n)=n-1,-1 \text{ (multiplicity } (n-1) \text{ )}\]
freckles
  • freckles
Now I know this isn't a method but it might help to come up with something from this if it is even true
freckles
  • freckles
like we know the trace of M is suppose to be the sum of the eigenvalues
freckles
  • freckles
\[0=\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5\]
freckles
  • freckles
I don't know a quick way to find the determinant but the product of the eigenvalues equals the determinant
zzr0ck3r
  • zzr0ck3r
ill play with this too
freckles
  • freckles
another cute thing \[M_n=[1]_n-I\] where 1_n is a n by n matrix only made of 1's and I is the identity matrix don't know if this helps yet
freckles
  • freckles
you know I'm not totally sure yet but it seems to be: in this case anyways \[\text{ eigenvalues(}M_n\text{)}=\text{ egivenvalues(}[1]_n\text{)}-\text{ egienvalues(}I\text{)}\]
freckles
  • freckles
the eigenvalues of I are 1 (n amount of times) and the eigenvalues of [1]_n are n, 0 (n-1 amount of times)

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